Physically,the self-inductance plays the role of

$A$ coil of self-inductance $L$ is connected in series with a bulb and an a.c. source. Brightness of the bulb decreases when

The back $e.m.f.$ induced in a coil,when current changes from $1 \ A$ to $0 \ A$ in $1 \ ms$,is $4 \ V$. The self-inductance of the coil is:

$A$ $10 \; \Omega, 20 \; mH$ coil carrying constant current is connected to a battery of $20 \; V$ through a switch. When the switch is opened, the current becomes zero in $100 \; \mu s$. The average emf induced in the coil is $\dots \; V$.

In the circuit shown below,a long time after the key $K$ is closed,the reading in the ammeter is $20 \,mA$. What was the reading (in $mA$) immediately after it was closed?

If a current of $10 \, A$ flows in one second through a coil,and the induced $e.m.f.$ is $10 \, V$,then the self-inductance of the coil is: